A large heavy box is sliding without friction down a smooth plane of inclination θ . From a point P on the bottom of the box, a particle is projected inside the box. The initial speed of the particle

Question

A large heavy box is sliding without friction down a smooth plane of inclination $θ$ . From a point $P$ on the bottom of the box, a particle is projected inside the box. The initial speed of the particle with respect to the box is $u$ and the direction of projection makes an angle $α$ with the bottom as shown in the figure:

Find the distance along the bottom of the box between the point of projection $P$ and the point $Q$ where the particle lands. [Assume that the particle does not hit any other surface of the box. Neglect air resistance]

TestHub · Accepted Answer

u is the relative velocity of the particle with respect to the box.

Resolve u.
$u_{x}$ is the relative velocity of the particle with respect to the box in x – direction.

$u_{y}$ is the relative velocity of the particle with respect to the box in y – direction.

Since there is no velocity of the box in the y – direction, therefore this is the vertical velocity of the particle with respect to ground also.

Y – direction motion (Taking relative terms w.r.t. box)

$u_{y} = + u \sin α$

$a_{y} = - g \cos θ$

$S_{y} = 0$ (activity is taken till the time the particle comes back to the box)

$t_{y} = t$

Using, $S = u t + \frac{1}{2} a t^{2}$

we get– $\begin{array}{l} 0 = (u \sin α) t - \frac{1}{2} g \cos θ \times t^{2} \\ \Rightarrow t = 0 o r t = \frac{2 u \sin α}{g \cos θ} \end{array}$

X-direction motion (Taking relative terms w.r.t. box)

$\begin{array}{l} u_{x} = + u \cos α \\ a_{x} = 0 \end{array}$

Again, using $S = u t + \frac{1}{2} a t^{2}$

we get – $\Rightarrow$ $S_{x} = u \cos α \times \frac{2 u \sin α}{g \cos θ} = \frac{u^{2} \sin 2 α}{g \cos θ}$

$[∵ t_{x} = \frac{2 u \cos α}{g \cos θ}]$

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